Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: jmxwell on October 14, 2020, 08:28:13 AM
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Hi all, my name is James. I'd like to know if someone can help me with this question:
Secondary alkyl halides preferentially react by SN2 mechanisms when exposed to strong nucleophiles and nonpolar solvents. However, in bicyclic secondary halides like the ones shown below, the SN2 reaction is about 100 times slower than the SN1 reaction under the conditions specified above. Would you be able to suggest a plausible explanation for this atypical behavior?
(https://i.ibb.co/Xy834Yd/23.jpg)
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What are your initial thoughts?
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I don't even can't start an explanation, I think this is because bond-making reactions
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It is a forum rule (see red link above) that you must provide your answer or at least some preliminary thoughts before we can help you. A good place to start might be to list out the factors that speed up or slow down SN2 reactions.
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...and what might speed up SN1 reactions....
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Well, I would explain this occurs because the leaving group, which will be responsible for best stability. That's why we have this : I -, Br- are better than Cl - and this is better than F -.
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What will form first before a Sn1 reaction? What is the geometry of this species?
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Bridgehead carbons generally have difficulties undergoing SN2 reactions because a bridgehead alkyl halide will be tertiary: bridgehead carbons are bonded to each of the two rings as well as the bridge between the two rings. Tertiary alkyl halides have a lot of difficulty undergoing SN2 reactions because the nucleophile would experience a lot of steric hindrance when trying to attack the alpha carbon. As a result, tertiary alkyl halides will generally undergo elimination reactions.
The greater the steric hindrance, the more difficult it will be for the nucleophile to attack.
A bridgehead carbon, however, will be especially unreactive to SN2 because of how SN2 proceeds via inverse of configuration. Inverting the stereochemical configuration of the bridgehead carbon will generate significant angle strain for the other carbons in the ring system, which especially for smaller cyclic compounds, will create a highly unstable product.
Yes?
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This is not a bridgehead halide. Check this link; https://en.wikipedia.org/wiki/2-Norbornyl_cation
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Are your example compounds tertiary alkyl halides?
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Secondary halides, I wrote wrong. But concept is ok?
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As Rolnor pointed out, these are not bridgehead alkyl halides. There may be more than one reason, but thinking about the geometry of the transition state of the SN2 reaction could help.
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Look at the link I gave, there is a special type of resonance-form for the carbeniumion.
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Well, I would explain this occurs because the leaving group, which will be responsible for best stability. That's why we have this : I -, Br- are better than Cl - and this is better than F -.
Damm even I am having a similar kind of issue, I have searched all over the internet and even have posted on number of threads on different forum, no solution seems to work. I am really frustrated, can anyone of you here help me resolve this issue, I am very much tired now.
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Can you formulate your question better?
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Thank you, everthing goes well under your explanation